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Exercise 9.6 · Q11

Q.Let the function ff be defined by f(x)={3x,0≤x≤1−3x+5,1<x≤2f(x)=\begin{cases}3x, & 0\le x\le1\\ -3x+5, & 1<x\le2\end{cases}, then

(1) lim⁡x→1f(x)=1\displaystyle\lim_{x\to1}f(x)=1
(2) lim⁡x→1f(x)=3\displaystyle\lim_{x\to1}f(x)=3
(3) lim⁡x→1f(x)=2\displaystyle\lim_{x\to1}f(x)=2
(4) $$\displaystyle\lim_{x\to1}f(x)doesnotexist does not exist
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Step 1. For x→1−x\to1^-, use the branch valid on [0,1][0,1]: f(x)=3x→3(1)=3f(x)=3x\to3(1)=3.

Step 2. For x→1+x\to1^+, use the branch valid on (1,2](1,2]: f(x)=−3x+5→−3(1)+5=2f(x)=-3x+5\to-3(1)+5=2.

Step 3. The left-hand limit (33) and right-hand limit (22) are unequal. …

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